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An irreversible engine operates between temperatures of 852 and 314 \(\mathrm{K}\) . It absorbs 1285 \(\mathrm{J}\) of heat from the hot reservoir and does 264 \(\mathrm{J}\) of work. \(\quad\) (a) What is the change \(\Delta S_{\text { universe }}\) in the entropy of the universe associated with the operation of this engine? \((\mathbf{b})\) If the engine were reversible, what would be the magnitude \(|W|\) of the work it would have done, assuming that it operated between the same temperatures and absorbed the same heat as the irreversible engine? (c) Using the results of parts (a) and (b), find the difference between the work produced by the reversible and irreversible engines.

Short Answer

Expert verified
(a) \( \Delta S_{universe} = 1.72 \, \text{J/K} \). (b) \( W_{rev} = 825 \, \text{J} \). (c) \( \Delta W = 561 \, \text{J} \).

Step by step solution

01

Calculate Entropy Change in the Hot Reservoir

For the irreversible engine, the change in entropy for the hot reservoir is calculated using the formula: \( \Delta S_{hot} = -\frac{Q_{in}}{T_{hot}} \), where \( Q_{in} = 1285 \, \text{J} \) and \( T_{hot} = 852 \, \text{K} \). Substituting the values, we have: \( \Delta S_{hot} = -\frac{1285}{852} \, \text{J/K} \).
02

Calculate Entropy Change in the Cold Reservoir

The change in entropy for the cold reservoir is calculated using the formula: \( \Delta S_{cold} = \frac{Q_{out}}{T_{cold}} \), where \( Q_{out} = Q_{in} - W = 1285 \, \text{J} - 264 \, \text{J} = 1021 \, \text{J} \) and \( T_{cold} = 314 \, \text{K} \). Substituting these values, we have: \( \Delta S_{cold} = \frac{1021}{314} \, \text{J/K} \).
03

Calculate the Change in Entropy of the Universe

The change in entropy of the universe, \( \Delta S_{universe} \), is the sum of changes in entropy of the hot and cold reservoirs: \( \Delta S_{universe} = \Delta S_{cold} + \Delta S_{hot} \). Add the results from Steps 1 and 2.
04

Calculate Work Done by a Reversible Engine

For a reversible engine, we use the Carnot efficiency: \( \eta_{rev} = 1 - \frac{T_{cold}}{T_{hot}} = 1 - \frac{314}{852} \). The work done \( W_{rev} \) is \( W_{rev} = \eta_{rev} \times Q_{in} \). Calculate it using \( Q_{in} = 1285 \, \text{J} \).
05

Calculate Difference Between Reversible and Irreversible Work

Subtract the work done by the irreversible engine from the work done by the reversible engine: \( \Delta W = W_{rev} - W_{irr} \), where \( W_{irr} = 264 \, \text{J} \). This will give the difference between the works.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy Change
Entropy change is a measure of disorder in a system. When dealing with heat engines, it’s important to calculate how entropy changes in both the system and its surroundings. For a heat engine, the focus is on how entropy changes in the hot and cold reservoirs.

- **Hot Reservoir**: Entropy decreases here because it transfers heat out to the engine. The change is negative as expressed through the formula \( \Delta S_\text{hot} = -\frac{Q_\text{in}}{T_\text{hot}} \). Using the given values, the calculation is \( \Delta S_\text{hot} = -\frac{1285 \text{ J}}{852 \text{ K}} \).
- **Cold Reservoir**: Here, entropy increases because it receives heat. The formula is \( \Delta S_\text{cold} = \frac{Q_\text{out}}{T_\text{cold}} \), and becomes \( \frac{1021 \text{ J}}{314 \text{ K}} \).

Adding these gives us the change in entropy for the universe related to this process, which is key in understanding the direction of natural processes.
Irreversible Engine
An irreversible engine does not operate at maximum theoretical efficiency due to factors like friction or thermal gradients. In this problem, the irreversible engine absorbs heat energy and does work, resulting in an increase in the total entropy of the universe. This is because the second law of thermodynamics dictates that real engines will always increase the universe’s entropy.

For the irreversible engine operating between the temperatures provided, the work is 264 J. This output highlights the actual energy conversion efficiency limits due to internal losses and disorder. Understanding the performance of this engine helps us appreciate why such engines cannot attain ideal performance.
Reversible Engine
A reversible engine operates in such a way that no energy is wasted or unordered. It works ideally, following the principles of thermodynamics, especially not increasing the universe's entropy in the process.

Such an engine is modeled using the Carnot cycle. A reversible engine calculates work using reversible efficiency: \( \eta_{\text{rev}} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \). Substituting the known temperatures, the efficiency comes out to be \( 1 - \frac{314}{852} \). Then, for work \( W_{\text{rev}} \), it is \( \eta_{\text{rev}} \times Q_{\text{in}} \). Calculating this gives an understanding of the ideal work possible, representing the benchmark efficiency for all heat engines.
Carnot Cycle
The Carnot cycle represents an idealized thermodynamic cycle proposed by Sadi Carnot focused on maximizing thermodynamic efficiency. This cycle is composed of two isothermal processes and two adiabatic processes.

- **Isothermal expansion**: The engine absorbs heat from the hot reservoir and performs work.
- **Adiabatic expansion**: The system does further work, causing it to cool without losing heat.
- **Isothermal compression**: Heat is expelled to the cold reservoir.
- **Adiabatic compression**: The system is compressed, heating back up to its original state.

Engines that ideally follow this cycle are considered 100% efficient in theory since they do not increase the total entropy of the universe. This underlines the core understanding of thermodynamic efficiency potential, serving as the maximum achievable efficiency for heat engines.

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Most popular questions from this chapter

(a) After 6.00 \(\mathrm{kg}\) of water at \(85.0^{\circ} \mathrm{C}\) is mixed in a perfect thermos with 3.00 \(\mathrm{kg}\) of ice at \(0.0^{\circ} \mathrm{C},\) the mixture is allowed to reach equilibrium. When heat is added to or removed from a solid or liquid of mass \(m\) and specific heat capacity \(c\) , the change in entropy can be shown to be \(\Delta S=m c \ln \left(T_{\mathrm{f}} / T_{\mathrm{i}}\right),\) where \(T_{\mathrm{i}}\) and \(T_{\mathrm{f}}\) are the initial and final Kelvin temperatures. Using this expression and the change in entropy for melting, find the change in entropy that occurs. \((\mathrm{b})\) Should the entropy of the universe increase or decrease as a result of the mixing process? Give your reasoning and state whether your answer in part (a) is consistent with your answer here.

A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignite the fuel occurs because the pistons compress the air in the cylinders. Suppose that air at an initial temperature of \(21^{\circ} \mathrm{C}\) is compressed adiabatically to a temperature of \(688^{\circ} \mathrm{C}\) . Assume the air to be an ideal gas for which \(\gamma=\frac{7}{5} .\) Find the compression ratio, which is the ratio of the initial volume to the final volume.

A system does \(4.8 \times 10^{4} \mathrm{J}\) of work , and \(7.6 \times 10^{4} \mathrm{J}\) of heat flows into the system during the process. Find the change in the internal energy of the system.

The inside of a Carnot refrigerator is maintained at a temperature of 277 \(\mathrm{K}\) , while the temperature in the kitchen is 299 \(\mathrm{K}\) . Using 2500 \(\mathrm{J}\) of work, how much heat can this refrigerator remove from its inside compartment?

The internal energy of a system changes because the system gains 165 \(\mathrm{J}\) of heat and performs 312 \(\mathrm{J}\) of work. In returning to its initial state, the system loses 114 \(\mathrm{J}\) of heat. During this return process, (a) what work is involved, and \((\mathrm{b})\) is the work done by the system or on the system?

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